Lecture 2'a: Linear Equations Method of Integrating Factors

1
Lecture 2.a Linear Equations Method of
Integrating Factors

• A linear first order ODE has the general form
• where f is linear in y. Examples include
  equations with constant coefficients, such as
  those in Chapter 1,
• or equations with variable coefficients

2
Constant Coefficient Case

• For a first order linear equation with constant
  coefficients,
• recall that we can use methods of calculus to
  solve

3
Variable Coefficient Case Method of
Integrating Factors

• We next consider linear first order ODEs with
  variable coefficients
• The method of integrating factors involves
  multiplying this equation by a function ?(t),
  chosen so that the resulting equation is easily
  integrated.

4
Example 1 Integrating Factor (1 of 2)

• Consider the following equation
• Multiplying both sides by ?(t), we obtain
• We will choose ?(t) so that left side is
  derivative of known quantity. Consider the
  following, and recall product rule
• Choose ?(t) so that

5
Example 1 General Solution (2 of 2)

• With ?(t) e2t, we solve the original equation
  as follows

6
Method of Integrating Factors Variable Right
Side

• In general, for variable right side g(t), the
  solution can be found as follows

7
Example 2 General Solution (1 of 2)

• We can solve the following equation
• using the formula derived on the previous slide
• Integrating by parts,
• Thus

8
Example 2 Graphs of Solutions (2 of 2)

• The graph on left shows direction field along
  with several integral curves.
• The graph on right shows several solutions, and a
  particular solution (in red) whose graph contains
  the point (0,50).

9
Example 3 General Solution (1 of 2)

• We can solve the following equation
• using the formula derived on previous slide
• Integrating by parts,
• Thus

10
Example 3 Graphs of Solutions (2 of 2)

• The graph on left shows direction field along
  with several integral curves.
• The graph on right shows several integral curves,
  and a particular solution (in red) whose initial
  point on y-axis separates solutions that grow
  large positively from those that grow large
  negatively as t ? ?.

11
Method of Integrating Factors for General First
Order Linear Equation

• Next, we consider the general first order linear
  equation
• Multiplying both sides by ?(t), we obtain
• Next, we want ?(t) such that ?'(t) p(t)?(t),
  from which it will follow that

12
Integrating Factor for General First Order
Linear Equation

• Thus we want to choose ?(t) such that ?'(t)
  p(t)?(t).
• Assuming ?(t) gt 0, it follows that
• Choosing k 0, we then have
• and note ?(t) gt 0 as desired.

13
Solution forGeneral First Order Linear Equation

• Thus we have the following
• Then

14
Example 4 General Solution (1 of 3)

• To solve the initial value problem
• first put into standard form
• Then
• and hence

15
Example 4 Particular Solution (2 of 3)

• Using the initial condition y(1) 2 and general
  solution
• it follows that
• or equivalently,

16
Example 4 Graphs of Solution (3 of 3)

• The graphs below show several integral curves for
  the differential equation, and a particular
  solution (in red) whose graph contains the
  initial point (1,2).

17
Lecture 2.b Separable Equations

• In this section we examine a subclass of linear
  and nonlinear first order equations. Consider the
  first order equation
• We can rewrite this in the form
• For example, let M(x,y) - f (x,y) and N (x,y)
  1. There may be other ways as well. In
  differential form,
• If M is a function of x only and N is a function
  of y only, then
• In this case, the equation is called separable.

18
Example 1 Solving a Separable Equation

• Solve the following first order nonlinear
  equation
• Separating variables, and using calculus, we
  obtain
• The equation above defines the solution y
  implicitly. A graph showing the direction field
  and implicit plots of several integral curves for
  the differential equation is given above.

19
Example 2 Implicit and Explicit Solutions (1
of 4)

• Solve the following first order nonlinear
  equation
• Separating variables and using calculus, we
  obtain
• The equation above defines the solution y
  implicitly. An explicit expression for the
  solution can be found in this case

20
Example 2 Initial Value Problem (2 of 4)

• Suppose we seek a solution satisfying y(0) -1.
  Using the implicit expression of y, we obtain
• Thus the implicit equation defining y is
• Using explicit expression of y,
• It follows that

21
Example 2 Initial Condition y(0) 3 (3 of 4)

• Note that if initial condition is y(0) 3, then
  we choose the positive sign, instead of negative
  sign, on square root term

22
Example 2 Domain (4 of 4)

• Thus the solutions to the initial value problem
• are given by
• From explicit representation of y, it follows
  that
• and hence domain of y is (-2, ?). Note x -2
  yields y 1, which makes denominator of dy/dx
  zero (vertical tangent).
• Conversely, domain of y can be estimated by
  locating vertical tangents on graph (useful for
  implicitly defined solutions).

23
Example 3 Implicit Solution of Initial Value
Problem (1 of 2)

• Consider the following initial value problem
• Separating variables and using calculus, we
  obtain
• Using the initial condition, it follows that

24
Example 3 Graph of Solutions (2 of 2)

• Thus
• The graph of this solution (black), along with
  the graphs of the direction field and several
  integral curves (blue) for this differential
  equation, is given below.